Chapter 12.4, Problem 32AYU

Extended Principle of Mathematical Induction The Extended Principle of Mathematical Induction states that if Conditions a and b hold, that is,

a. A statement is true for a natural number $j$.

b. If the statement is true for some natural number $k\ge j$, then it is also true for the next natural number $k+1$.

Then the statement is true for all natural numbers $\ge j$. Use the Extended Principle of Mathematical Induction to show that the number of diagonals in a convex polygon of $n$ sides is $\frac{1}{2}n\left(n-3\right)$.

[Hint: Begin by showing that the result is true when $n\text{ = }4$ (Condition (a).]

To determine

To prove: The statement

‘The number of diagonals in a convex polygon of $n$ sides is $\frac{1}{2}n\left(n - 3\right)$’ is true for all natural numbers $n$ using the Extended Principle of Mathematical Induction.

Answer to Problem 32AYU

The statement is true for the natural number $\left($ k + 1$\right)$ terms, it is true for all natural numbers $\ge j$ by the extended principle of mathematical induction.

Explanation of Solution

Given:

Statements says

‘The number of diagonals in a convex polygon of $n$ sides is $\frac{1}{2}n\left(n - 3\right)$’ is true for all natural number.

Formula used:

The Extended Principle of Mathematical Induction

Suppose that the following two conditions are satisfied with regard to a statement about natural numbers:

CONDITION I: A statement is true for a natural number $j$.

CONDITION II: If the statement is true for some natural number $k \ge j$, it is also true for the next natural number $k + 1$. Then the statement is true for all natural numbers $\ge j$.

Proof:

Consider the statement

‘The number of diagonals in a convex polygon of $n$ sides is $\frac{1}{2}n\left(n - 3\right)$’    -----(1)

Step 1: Show that statement (1) is true for the natural number $j = 4$.

Number of sides = 4. 4-sided polygon is nothing but quadrilateral. In general, quadrilateral has two diagonals. Hence the statement is true for the natural number $j = 4$.

Step 2: Assume that the statement is true for some natural number $k$.

That is ‘The number of diagonals in a convex polygon of $k$ sides is $\frac{1}{2}n\left(n - 3\right)$’    -----(2)

Step 3: Prove that the statement is true for the next natural number $k + 1$.

‘The number of diagonals in a convex polygon of $(k + 1)$ sides is $\frac{1}{2}\left(k + 1\right)\left(k + 1 - 3\right) = \frac{1}{2}\left(k + 1\right)\left(k - 2\right) = \frac{1}{2}\left(k^2 - k - 2\right)$’

For a convex n-sided polygon, there are n vertices and from each vertex you can draw $n - 3$ (we cannot draw diagonals to its own, the predecessor and successor vertices) diagonals.

From equation (2), it can be concluded that for $k$ vertices we can draw $\frac{1}{2}k\left(k - 3\right)$ diagonals

For $n = 5$, extra 3 Diagonals are added to $n = 4$ and $\text{Number}\text{of}\text{diagonals} = 2 + 3 = 5$,

For $n = 6$, 4 extra Diagonals added to $n = 5$ and $\text{Number}\text{of}\text{diagonals} = 5 + 4 = 9$,

For $n = 7$, 5 extra Diagonals added to $n = 6$ and $\text{Number}\text{of}\text{diagonals} = 9 + 5 = 14$,

For $n = 8$, 6 extra Diagonals added to $n = 7$ and $\text{Number}\text{of}\text{diagonals} = 14 + 6 = 20$.

In general, for $k + 1$ side polygon, $\left(k + 1\right) - 2$ extra diagonals will added to $k$ side polygons.

So the number of diagonals in a convex polygon of $\left($ k + 1$\right)$ sides is $\frac{1}{2}k\left(k - 3\right) + \left(k + 1\right) - 2$

$= \frac{1}{2}k\left(k - 3\right) + k - 1 = \frac{1}{2}\left(k^2 - 3k + 2k - 2\right) = \frac{1}{2}\left(k^2 - k - 2\right)$

As the statement is true for the natural number $\left($ k + 1$\right)$ terms, it is true for all natural numbers $\ge j$ by the extended principle of mathematical induction.